HDU 3836 —— Equivalent Sets

来源：互联网发布：人脸识别asm算法编辑：程序博客网时间：2024/06/04 20:00

原题：http://acm.hdu.edu.cn/showproblem.php?pid=3836

题意：问至少要加几条边，使得该图强连通（即图中任意两点都连通）；

思路：先求原图的强连通分量，然后进行缩点构建新图，在新图中，求入度 = 0 和出度 = 0 的个数，两者的最大值即为答案；

#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;const int maxn = 200000+5;const int maxm = 500000+5;int n, m;int head[maxn], edgenum;int Stack[maxn], Belong[maxn], DFN[maxn], Low[maxn];int Time, taj, top;bool Instack[maxn];int in[maxn], out[maxn];struct Edge{int from, to;int next;}edge[maxm];void Tarjan(int u){DFN[u] = Low[u] = ++Time;Stack[top++] = u;Instack[u] = true;for(int i = head[u];i != -1;i = edge[i].next){int v = edge[i].to;if(DFN[v] == -1){Tarjan(v);Low[u] = min(Low[u], Low[v]);}else{if(Instack[v] && Low[u] > DFN[v])Low[u] = DFN[v];}}if(DFN[u] == Low[u]){taj++;while(1){int now = Stack[--top];Belong[now] = taj;Instack[now] = false;if(now == u)break;}}}void add(int u, int v){edge[edgenum].from = u;edge[edgenum].to = v;edge[edgenum].next = head[u];head[u] = edgenum++;}void init(){memset(head, -1, sizeof head);edgenum = 0;memset(DFN, -1, sizeof DFN);memset(Instack, false, sizeof Instack);Time = taj = top = 0;}int main(){while(~scanf("%d%d", &n, &m)){init();while(m--){int u, v;scanf("%d%d", &u, &v);add(u, v);}for(int i = 1;i<=n;i++){if(DFN[i] == -1)Tarjan(i);}if(taj == 1){printf("0\n");continue;}memset(in, 0, sizeof in);memset(out, 0, sizeof out);for(int i = 0;i<edgenum;i++){int u = Belong[edge[i].from];int v = Belong[edge[i].to];if(u != v){out[u]++;in[v]++;}}int incnt = 0, outcnt = 0;for(int i = 1;i<=taj;i++){if(in[i] == 0)incnt++;if(out[i] == 0)outcnt++;}printf("%d\n", max(incnt, outcnt));}return 0;}

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